Help in Chem: Balancing Oxidation-reduction equations

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1

<p>Thanks, I'm trying to learn how to do it because I don't have a firm grasp on the subject</p>

<p>Anyway, suppose we have </p>

<p>K2Cr2O7 + HCl ==> KCl + CrCl3 + H2O + Cl2</p>

<p>So we do the oxidation half-reaction</p>

<p>Cr2O72- ==> Cr3+</p>

<p>Cr2O72- ==> 2 Cr3+</p>

<p>Cr2O72- ==> 2 Cr3+ + 7 H2O</p>

<p>14 H+ + Cr2O72- ==> 2 Cr3+ + 7 H2O</p>

<p>14 H+ + Cr[sub]2[/sub]O72- + 6 e- ==> 2 Cr3+ + 7 H2O</p>

<p>My question is where did the 6e- come from?</p>

2

<p>well the total charges on the left have to equal to the total charges on the right</p>

<p>so on the left we has +12(14-2) and on the right we have +6(2*+3). So in order to balance to the total equation we need to add 6 electrons to the left to make the left have a charge of +6.</p>

3

<p>Oh I see, so balance Cr2 -> 2Cr first before balancing the electrons. Thank you so much I understand now :)</p>